The bond angle of water is decreased from 109Degree 28' to 104Degree 24',why?

Answer 1

The two sets of unpaired electrons on the water molecule repel each other making the H-O-H angle less than the 108.29 theoretical value.

Answer 2

The valence shell of Carbon (forming CH4), Nitrogen (forming :NH3) and Oxygen (forming HOH with 2 non-bonded pair) undergoes hybridization into 4sp^3 Hybrid Orbitals which look like skewed figure 8s... This is like taking the valence shell orbitals and putting them all in a blender, mixing really well and pouring the results out into 4 equal piles. These are hybrid orbitals. They all look alike, all have same energy and each can hold up to two electrons. Now, just before bonding takes place, the 4 hybrid orbitals connect at their apex of the fig-8 and assume positions in space of a regular tetrahedron. For CH4 all bonds are equal and angles between hybrids = 109.50 degrees. For Ammonia (:NH3) one of the hybrids is carrying 1 non-bonded pair of electrons and the other hybrids have 1 paramagnetic (unpaired) electron which bonds with Hydrogen. The negative character of the non-bonded pair has a stronger negative character than the bonded electron pairs. This results in e/e repulsions to push the bond angels to 107-degress and Oxygen in H:O:H (Visualize two non-bonded pair above and below the oxygen.) The stong negative character of the 2 non-bonded electron pairs repel the bonded electron pairs to 105-degrees.

Note that substrates that have high electronegative character (Halogens) attached to one of the hybrid orbitals of carbon can function to attract the bonded electron pair and increase the bond angles to greater than 109-degrees... Example, CH3Cl (Methy Chloride).

For more info... Google Valence Bond Theory of Molecular Bonding.